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LIBRARY OF CONGRESS. 
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UNITED STATES OF AMERICA. 



THEORY OF TURBINES. 



DE YOLSON WOOD, 

Professor of Mechanical Engineering, Stevens Institute of Technology. 






J. WILEY & SONS, 
NEW YORK. 

18 9 5. 



Copyright, 1893, by 
DE VOLSOST WOOD. 






/ 



Remark. 

About forty-live years ago M. Poncelet made a solution of the 
Fourneyron turbine which, for its thoroughness and the direct- 

Dess of its analysis, has become classical (Comptes jRendus, 1838). 
But that writer neglected the frictional (and other) resistances 
within the wheel, and assumed that the buckets, or passages in 
the wheel, were constantly full. The former is an important 
element in the theory, and its consideration makes the analysis 
but little more complicated. 

AVeisbach, in his Hydraulic Motors, gives a solution in which 
frictional resistances are involved, and the sections of the stream 
at the outlet of the supply chamber, the entrance into the wheel, 
and all the sections of the buckets are determined when the 
wheel runs for best efficiency. The formulas, however, are so 
complex that but little practical knowledge can be gained from 
•their general discussion. I have, therefore, assumed that the 
wheels here discussed have about the proportions made for com- 
mercial purposes, and deduced certain numerical results which 
are entered in tables ; and a simple examination of these furnishes 
certain desirable information. 

The driving power here considered is that of an incompressible 
fluid, which in practice will be water. The steam turbine or 
those driven by a compressible fluid are not in practice con- 
structed like water turbines, and no theory for such is here at- 
tempted. 



HYDRAULIC MOTORS. 



1. The motors here analyzed may be called "reaction 
wheels," or "pressure turbines." Some writers call those 
wheels " pressure turbines " in which the water has a "free 
surface," and these by others are called " turbines of free devi- 
ation. (Weisbach, Hydraulics, etc., p. 421.) The former term 
is somewhat ambiguous when applied to wheels in which the 
water in the buckets has a free surface, and, therefore, the 
term " free deviation " will be applied to such. 

There will first be given a general solution of the " pressure 
turbine," and the other turbines will be considered as special 
cases of the more general one. 

2. Notation. 

Let Q be the volume of water passing through the wheel 
per second, 

6, the weight of unity of volume of the water, or 62 J pounds 
per cubic foot ; then 

6Q=^Y will be the weight of water passing through the 
wheel per second, 

li x be the head in the supjDly chamber above the entrance 
to the buckets, 

lu, the head in the tail race above the exit from the bucket, 

2,, the fall in passing through the buckets, 

H—h { ^ r z l — Jt 2 , the effective head, 

U, the useful work done by the water upon the wheel, 

R, the work lost by frictional resistances, whirls, etc., 

Mi, the coefficient of resistance along the guides, 

M2, the coefficient of resistance along the buckets, 



HYDRAULIC MOTORS. 



r lt the radius of the initial rim, 
r 2 , the radius of the terminal rim, 
p, the radius to any point of the bucket, 
n=7\-^-r 2 , the ratio of the initial to that of the terminal radius, 
V, the velocity of the water issuing from supply chamber, 
Vi, the initial velocity of the water in the bucket in reference 

to the bucket, 
<oj- v, the velocity 

along the bucket 
at any point, 

v 2 , the termi- 
nal velocity in 
the bucket, 

ii/",theveloc-ity 
of exit in refer- 
ence to the earth, 
(<o, the angular 
velocity of the 
wheel, 

<x 9 terminal 
angle between 
the guide and in- 
itial rim — CA B, 
~/ u angle between the initial element of bucket and initial 
rim = EAD, 

y 2 = GFI, the angle between the terminal rim and terminal 
element of the bucket, 

6 — EFL angle between the terminal rim and actual direc- 
tion of the water at exit, 

p lt the pressure of water at entrance of the bucket per unit 
area, 

p, the pressure of water at any point of the bucket, 

p iy the pressure of water at exit, 

p a , the pressure of an atmosphere, 

a — eb, the arc subtending one gate opening, Fig. 3, 




Fig. 1. 



HYDRAULIC MOTORS. 6 

«,, the arc subtending one bucket at entrance. (In Fig. 3, 
a and Oh appear to be the same but in practice they are usually 
different, a being greater than «,.) 

Oj = gh, the arc subtending one bucket at exit, 

K — bf\ normal section of passage, it being assumed that 
the passages and buckets are very narrow, 

k { = bd, initial normal section of bucket, 

k, = gi t terminal normal section, 

Y, the depth of K, y % of k h and ?/ 2 of k . 
Then 



K = Ya sin a ; 

A'i = y [ a l sin y, 
l\ 2 = y. 2 a 2 sin 2 



(1) 




Fig. 2. 



ojTi = velocity of initial 
rim, 

cjr 2 = velocity of terminal 
rim. 

3. General Solution. 

Beginning with the pres- 
sure on the top of the supply 
chamber, the relation between 
the heads, actual and virtual, will be determined to the point 
of discharge from the wheel. 

The pressure per unit on the upper surface of the supply 
chamber will be that of the atmosphere, or 

Pa, 

and the corresponding virtual head in terms of a column of 
water will be 

Pa 



The head in the supply chamber above the entrance to the 
wheel will be 



HYDRAULIC MOTORS. 



therefore, the total head above the initial element of the bucket 
will be 

This head produces an actual pressure p x at the entrance 
to the bucket and the velocity V of exit from the guides ; 
hence, according to Bernoulli's theorem, the heads due to the 
pressure p x and velocity V\ will equal the former, or 

P 

(2) 



\ 



6 $ 



Pl=Pa+ ^l 



e P 



(3) 




which will be the theoretical pres- 
sure at entrance to the bucket if 
friction be neglected. Represent 
the head lost by friction by 

T- 



l( i 



w 



which must also be overcome by 
the head in the supply chamber, 
so that we have, by adding it to 
the second member of (2) and transforming, 



(1 + mi) V 2 = 2g] h + 2g 



Pa - P\ 



The triangle of velocities ABC, Fig. 1, gives 

JZ sin Y\ sin y l 

sm (« + yj sin (a + y Y ) 



sin a 



GOT 



w 



(5) 



(6) 



sin (a + y x ) 
The relation between the initial and terminal velocities in 



HYDRAULIC MOTORS. 



the bucket involves the velocity of the wheel and the pressure 
in the bucket. 

Let m be an elementary mass at a distance p from the axis 
of the wheel, then will the centrifugal force be 




Fig. 4. 



and if this element by moving a distance ds in the tube also 
moves outward a distance dp, the work done by the centrifugal 

force will be 

moj 2 pdp. 

If the tube (or bucket) be 
inclined downward, the work 
done i or energy acquired) by 
the weight in falling a distance 
dz will be 

mgdz. 

These two works will be ex- 
pended in the following ways : 

a. Increasing the energy of the water in the tube in refer- 
ence to the tube by an amount 

\ md (v 2 ). 

b. In doing work against the difference of pressures on the 
two faces of the element, and considering the back pressure 
p greater than the forward pressure p', the work will be 

dp 

™g ~\ 

where dp -r S is equivalent to a head through which mg would 
work. 

c. In overcoming frictional resistance. The law of fric- 
tional resistances is not well known, but is assumed to vary 
as the energy of the mass and wetted perimeter. The pe- 
rimeter is here discarded, hence the work will be 



b HYDRAULIC MOTORS. 

/*„ • \ mv 2 ds. 
Hence we have 

mgdz + moo^pdp — \ md (v 2 ) + mg -— + -k jJ 2 mv 2 ds, ... (7) 

But the last term cannot be integrated unless v be a 
known function of s, and since this is not known, we make v = 
v 2 , the terminal velocity. The coefficient /a 2 is determined 
independently of the length, and includes the value yu 2 -s, when 
s is the length of a bucket. 

Integrating between initial and terminal limits gives 

Go 2 (r 2 2 - r>) _ vj - v x 2 p-2-Pi v 2 2 , . 

Zl + ' Yg 2^" + ~^ +/ * 2 %" * (8) 

The fall z y is so small in practice compared with the next 
term of the equation, that it may, and will, be omitted, giving 

(9) 



(1 + M 2 ) v} = v? + erf (r 2 2 — r, 2 ) - 


_O a P2~Pl 


• 


which gives v 2 . 

At exit the pressure will be 






P-2=Pa + $K • 







(10) 

The velocity of exit, relative to the earth, will be 

w 2 = 1) 2 2 + G0> V — 2v 2 .^^ 2 cos^ 2 (11) 

The work done upon the wheel will be the initial (poten- 
tial) energy of the water less the energy in the water as it 
quits the wheel, still further diminished by the energy due to 
frictional losses ; or 

U=dQH-6Q~-B (12) 

/? = ^-£ + MG |° (13) 



HYDRAULIC MOTORS. 



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HYDRAULIC MOTORS. 



= L [— JfV 2 oti + cos ;/ 2 V2gH + iWgca 3 ] r 2 &?, 
when 

Z= l 



(15) 



gllVl + av 2 



■M 2 = Vl~+7<i 



cos a' sin yifi 



sin i « + Xv \ r 



JV' = 1 



2 cos a sin ^ sin 



+ / 



sm(ar+ Y\) sin J (/* + y^y \r 2 



(15a) 



For maximum efficiency make dE ■+■ dco = in (15) and solve 
for oo, calling this particular value co\ then 



CsD — 



\/a#. /Jf'-VJf 4 - 2P 



'^ iTVJf 4 



COS" ^2 



V J/ 4 — iV" QOsVa 



(16) 



which value substituted in equation (15) will give the maxi- 
mum efficiency. Then equations (5), (6>, become 



-,-t sin v\ 

V = -t— ; " r^/ 1 ], 



sin (or 4- i/ T ) 
sin a 



v i = zr. 



sin (a + ;/ 2 ) 
Also from (9), (10), (4), (17), (18 



GOT 



(17) 
(18) 



But 



siircr — ^in 2 ^) 



Sid- a — sin'-/, 
sin 2 (a + Y\) — sin 2 a — sin-^i -+- 2 sin (a + y ^ cos a sin y 

2 sin (a + y^) cos a sin jk, 

— ~~ Am a + y ^ sin (a — y ^) + 2 sin(a + ^])cos a siny 1 

2 cori a sin v , 
sin (a + Y\) ' 

which substituted above will give equation (14)= 



HYDRAULIC MOTORS. 



V 2 



v& 

a /o tt . / 2 IcosTTsin Ki 2 sin 2 ;/! 2 \ « m 

y y \- sm (« +y,) sin 8 (a + n) V v ; 

The normal sections of the buckets will be 

K=$; h = Q-; h=^\ *=2. . . (20) 
I ^ y 2 v 

The depths of those sections will be 

Y= J£— 9 y x = ft , ^ = A ' 2 , . (21) 
i sib « * a x sin yi a 2 siny 2 



DISCUSSION. 

4. Three simple systems are recognized. 

r x < r 2 , called outivard flow. 

it i > T 2i called inward flow. 

r t = r 2 , called parallel floiv. 
The first and second may be combined with the third, mak- 
ing a mixed system. The third, in theory, is really an inward 
or outward flow, with an indefinitely narrow crown, although 
the analysis applies to a parallel flow wheel, in which the 
width is indefinitely small, and depth small compared with 
the total head. 

5. Value of y-i, the quitting angle. 

Equations (14) and (15) show that the efficiency is increased 
as cos y 2 is increased, or as y 2 decreases, and is greatest for ;/ 2 =0. 
Hence, theoretically, the terminal element of the bucket should 
be tangent to the quitting rim for best efficiency. This, how- 
ever, for the discharge of a finite quantity of water, would re- 
quire an infinite depth of bucket, as shown by the third of equa- 
tions (21). In practice, therefore, this angle must have a finite 



10 HYDRAULIC MOTORS. 

value The larger the diameter of the terminal rim the smaller 
may be this angle for a given depth of wheel and given quantity 
of water discharged. Theoretical considerations then would 
require, for best efficiency, a very large diameter for the quit- 
ting rim, and a very small angle, y 2 , between the terminal ele- 
ment of the bucket and the rim ; but commercial considera- 
tions require some sacrifice of best efficiency to cost, so that 
a smaller diameter and larger angle of discharge is made. If 
wheels are of the same diameter and depth, the inward flow 
wheel requires a larger quitting angle for the same volume 
of water than the outward now, since the discharge rim will be 
smaller in the former tha'n in the latter wheel, and the velocity 
v 2 , eq. '19), will also be less. In practice y 2 is from 10° to 20°. 

6. Relation betiveen y 2 and go'. 

Equation (16) when put under the form 



, VgH A / 1 

m = ~kV ,/ T^^T. - 1 - (22) 

shows that <*> increases as y 2 decreases, and is largest for 
y 2 — ; that is, in a wheel in which all the elements except y 2 
are fixed, the velocity of the wheel for best effect must increase as 
the quitting angle of the bucket decreases. 

If the terminal element be radial, then y 2 = 90°, and equa- 
tion (22) gives go' = ; that is, for minimum efficiency the 
wheel must be at rest, and no work will be done. 

7. Values of a + y v 

If a + y t = 180°, and a and y l both finite, then will M and 
N in (15a) both be infinite ; but equation (5) gives 

sin 180° V A /00x 

go — — : ■ — = ; (23) 

sm y x 7\ J 

that is, the wheel will have no motion, and no work will be 



HYDRAULIC MOTORS. 1 1 

done. If a -f- y t = 180°, then the terminal element of the 
guide and' the initial element of the bucket have a common 
tangent, in which case the stream can flow smoothly from the 
former into the latter only when the wheel is at rest. (See 
Fig. 5.) 

If a + y l exceed 180°, go' would be negative, and it would 
be necessary to rotate the wheel backwards in order that the 
water should flow smoothly from the guide into the bucket. 

It follows, then, that a + y x must be less than 180°, but the 
best relation cannot be determined by analysis ; however, 
since the water should be deflected from its course as much as 
possible from its entering to its leaving the wheel, the angle a 
for this reason should be as small as practicable. 

8. Values of a. 

If at = 0, equation i!4) will reduce to 



E = 



(/HVl + M-2 



VI + /< 2 (- r} + r, 2 ) go 



■ 2 cos y 2 V 'IgH + (r 2 2 — 2r L 2 — Min 2 ) go 2 



(24) 



which is independent of y Y ; hence, for this limiting case, the 
efficiency will be independent of the initial angle of the bucket. 
This is because the water enters the wheel tangentially and 
therefore has no radial component that would give an initial 
velocity in the bucket ; and equation (18) shows that the ini- 
tial velocity v x would be zero, while (17) shows that the velocity 
of the initial rim must equal that of the water flowing from the 
guides, or 

For the limiting, or critical case, 

a — 0, y 2 = 0, u x = 0, m 2 = 0, 

the velocity producing maximum efficiency will be, from equa- 
tion (16), 

r^' = Vg~H~ (25) 



12 



HYDRAULIC .MOTORS. 



or the velocity of the initial rim, if the wheel be frictionless, 
will be that due to half the head in the supply chamber. 
If r£ = 2?y, then 

r 2 GD' = V2gH, (26) 

or the velocity of the terminal rim will equal that due to the 
head. Substituting in (19) the values a — 0, y 2 = 0, u x = 0, 
u. 2 = 0, r} = 2?*! 2 , and it will reduce to 

v 2 = V2gH, (27) 

as it should. 

The following table gives the values of quantities for the 
three classes of wheels :* 

TABLE I. 



a = 0, 



y-i =0, 



u l = 0. 



JU-2 = 0. 





Velocity of 


Velocity 

of Exit 

from 

Guide 

V. 


Velocity 

Initial 


in Bucket. 

Terminal 
v 9 . 


Velocity 

of Exit. 
w. 


Efficiency, 
E. 


Dimensions 

of Wheel. 


Inner Rim. 


Outer 

Rim. 




co >, 


go r 2 












r l = ^/ir, 


y/gfi 


^/2g~H 


Vg~s 


0.00 


y/tgii 


0.00 


1.000 


r ] =r. 2 


go r. 2 


Vgs 


y/W 


0.00 


\/gM 


0.00 


1.000 


r { =1.4r 2 


0.7lWgH 


\ ffH 


y/g~S 


0.00 


lU^/jH 


0.00 


1.000 



In the first case the inner rim is the initial one, in the 
third case the outer rim is initial, it being an inward flow 
wheel. 

Since, in this case, the velocity of admission to the wheel 
in reference to the earth is that due to half the head in the 
supply chamber, and the velocity of exit is zero, it follows 
that the energy due to the velocity is all imparted to the 



HYDRAULIC MOTORS. 13 

wheel ; and the energy clue to the remaining half of the head 
is imparted to the wheel by pressure in the wheel. If the 
velocity of entrance to the wheel be that due to the head, or 
2 ill then will no energy be imparted to the wheel on 
account of pressure exerted by any part of the head //, but if 
V- < 2 ///, then will some of the work be done by this press- 
ure, w being zero. For the cases in Table I., the energy im- 
parted to the wheel will be due one-half to velocity and one-half to 
pressure : or in symbols, 

.w 



~>~ 



. gH + ±WH= Wff 9 . . (28) 



or, the entire potential energy of the water will be exjoended 
in work upon the wheel. 

Whenever V' < ZgH, the pressure at entrance must exceed 
the external pressure at exit, and 



H 



(29) 



then will be the part of the head producing pressure in the 
wheel. 

In practice, ol cannot be zero and is made from 20° to 30°. 
When other elements of the wheel are fixed, the value of a 
may be determined so as to secure a certain amount of initial 
pressure in the wheel, as will be shown hereafter. 

The value )\ — 1Ai\ makes the width of the crown for 
internal flow about the same as for r x — ^/lr 2 for outward flow, 
being approximately 0.3 of the external radius. 

9. Values of )* x and pt .' 

The frictional resistances depend not only upon the con- 
struction of the wheel as to smoothness of the surfaces, sharp- 



14 



HYDRAULIC MOTOES. 



ness of the angles, regularity of the curved parts, but also 
upon the manner it is run ; for if run too fast, the initial ele- 
ments of the wheel will cut across the stream of water, pro- 
ducing eddies and preventing the buckets from being filled, 
and if run too slow, eddies and whirls may be produced and 
thus the effective sections be reduced. These values cannot 
be definitely assigned beforehand, but Weisbach gives for 
good conditions, 

IA X = /.i 2 = 0.05 to 0.10 (30) 

They are not necessarily equal, and /*, may be from 0.05 to 
0.075, and m 2 from 0.06 to 0.10, or values near these. 

10. Values of y x . 

It has already been shown that y 1 must be less than 
180° — a. If y x — 90°, equation (14) shows that the efficiency 
of the Motionless wheel will be independent of a. The effect 
of different values for y x is best observed from numerical 
results as shown in the following table : 

TABLE II. 



Let a = 25°, 



y 2 = 12°, 



0.10. 



Initial 

Angle. 

Yi- 

(1) 


r } 


- /•„ >/} 




/ 


i = \Ar 


!• 


(2) 


E. 
(3) 


(4) 


<o'r 2 . 

(5) 


E. 
(6) 


(7) 


60° 


1.322vW 


.812 


.984V^ff 


.780^^ 


.911 


1.092-v/^ff 


90° 


1.226 " 


.827 


.866 " 


.689 " 


.908 


.964 " 


120° 


1.078 " 


.838 


.762 " 


.576 " 


.898 


.806 " 


150° 


.518 " 


.744 


.366 " 


.271 " 


.752 


.379 " 



The values t»'r a in columns (2) and (5) are velocities for the 
terminal rim, which in column (2) are for the exterior rim, but 



!IYI>1;.YU,1C MOTORS. 



15 



for column (5) it is the interior rim, while column (7) is for 

the exterior rim. 

Columns (2) and (7) show that the velocity of the outer 
rim is loss, for maximum effect, for the inflow than for the out- 
How, for the same size wheel. 

Column (3) shows that the efficiency, £] decreases as the 
initial angle of the bucket, y v increases up to 120°. This 
maximum will be for this wheel with this amount of friction. 

Column (6) shows that for the inflow wheel the efficiency 
continually decreases as y i increases. If the head and quan- 
tity of water discharged be constant, the work would be pro- 
portional to the efficiency ; for, from equation ( 14\ 



The effect of 



U= SQHE ....... (31) 

on the velocities is shown in Table III. 



TABLE III. 



Let a = 25 c 



r-> 



12\ 



iio =0.10, 



Ini- 


r x .-= r^/h 


/' 1= 1.4r 2 . 


tial 
Angle. 


V 


VgH 
.396 


«2 


K x fc, x 
VgH Vf/H 

1.219 2.525 


* 2 * 
VgH 

.691 


F 

VgH 

.959 


VgH 
.463 


V2 

VgH 
.761 


A'x 

VgH 


*, X 

VgH 
2.100 


A- 2 x 




VgH 
.820 


VgH 
1.447 


VgH 


60 3 


1.043 


1.314 


90° 


.955 


.403 


1.378 


1.0472.481 


.725 


1.063 


.449 


.676 


.940 


2.227 


1.479 


120° 


1.150 


.560 


1.153 


.869 1.785 


.874 


1.217 


.593 


.605 


.821 


1.686 


1.653 


150° 


2.100 


1 . 775 


.621 


.476 .568 


1.610 


2.060 


1.741 


.296 


.485 


.574 


3.378 



For commercial considerations it may be necessary to sac- 
rifice some efficiency to save on first cost, and to avoid making 
the wheel unwieldy. 

From equation (4) it appears that the pressure in the wheel 
at entrance, p u diminishes as the velocity of admission, T , in- 



16 HYDRAULIC MOTORS. 

creases, and. according to equation (5), V depends upon y t 
when a is fixed. Since the crowns are not fitted air tight nor 
water tight it is desirable that j^ should exceed the pressure 
of the atmosphere when the wheel runs in free air, or the press- 
ure p> + Pa when submerged, to prevent air or water from flow- 
ins in at the edge of the crown. It will be shown hereafter, in 
discussing the pressures in the wheel, that we should have 





— 


tan y l > tan 2^, . 


or, 


180 


-ri>**, 


or, 




;/, < 180 c - 2« 


If 




- a = 30, 


then 




n < 120°, 



(32) 



To be on the safe side, the angle y l may be 20 or 30 degrees 
less than this limit, giving 

n = 180° - 2a - 25 (say). 
= 155 - 2a. 

Then if a = 30°, y x = 95°. Some designers make this angle 
90°, others more, and still others less than that amount. Weis- 
bach suggests that it be less so that the bucket will be shorter 
and friction less. This reasoning appears to be correct for the 
inflow wheel, for the size and conditions shown in Table II., 
but not for the outflow wj.eel. In the Tremont turbines, de- 
scribed in the Lowell Hydraulic Experiments, this angle is 90°, 
the angle a, 20", and y%, 10°. Fourneyron made y 1 = 90°, and 
a from 30° to 33 . 

In Table III. it appears that for y x = 150 = , V = 2.1 Vgft, 
which exceeds V \gH; that is, the velocity of exit from the 
supply chamber exceeds that due to the head, which condi- 
tion must result from a negative pressure at entrance into the 
wheel. For zero pressure for the frictionless wheel, the above 
condition gives 

y x = 180° - 2 a, 



HYDRAULIC MOTORS. 

which for <> = 2d , gives )\ = 130 , and for v x = 150", the press- 
ure would be negative, and for 120 it would be positive. It 
appears that for the wheel with friction, considered in the table, 
that this pressure is also positive for y i = 1*20', and negative 
for 150 . 

11. Form of Bucket. 

The form of the bucket does not enter the analysis, and 
therefore its proper form cannot be determined analytically. 
Only the initial and terminal directions enter directly, and from 
these and the volume of the water flowing through the wheel, 
the area of the normal sections may be found from equations 
(20). 

But well-known physical facts determine that the changes of 
curvature and section must be gradual, and the general form 
regular, so that eddies and whirls shall not be formed. For the 
same reason the wheel must be run with the correct velocity 
to secure the best effect ; for otherwise the effective angles a 
and y x may be changed to values which cannot be determined 
beforehand, in which case the wheel cannot be correctly ana- 





Fig. 5. Fig. 6. 

lyzed. In practice the buckets are made of two or three arcs 
of circles mutually tangential at their points of meeting. Also, 
if the normal sections, K, \, h, of the buckets as constructed 
do not agree with, those given by computation, the stream will, 
if possible, adjust itself to true conditions by the formation of 



18 



HYDRAULIC MO TOES. 



eddies. If the teiminal sections at the guides, or the initial 
section of the bucket, be too small, the action may be changed 
from a pressure wheel to one of free deviation. So long as the 
pressure in the wheel exceeds the external pressure, the pre- 
ceding analysis is applicable for the wheel running for best 
effect, observing that the sections K, k l9 &,, are not those of 
the wheel, but those which are computed from the velocities 



V, v„ 



i\. 



12. Value of ; or direction of the quitting water. 
From Fig. 1 it may be found that 



and 



w cos = v 2 cos y 
to sin 6 = v 2 sin y 

.'. cot 6 = cot y 2 - 



got* 



go r 9 



v 2 sin y 2 



(33) 
(34) 

(35) 



These formulas are for the velocity giving maximum effi- 
ciency. If the speed be assumed, go in place of go' becoming 
known, v 2 is given by equation (19). It is apparent for such a 
case that 6 may have a large range of values from 6 = y 2} when 
the wheel is at rest, to 6 exceeding 90° for high velocities. The 
following table gives some results : 

TABLE IV. 



ar=25 C 



r- 2 = i2 c 



/J, = jli 2 = 0.10. 





?' 1 = 


r* Vh 


r,= 


1.4 r a 


Vi 


CO 


e 


CO 





60° 


.314 V^H 


72° 14' 


.160 VgR 


102° 43' 


90° 


.310 " 


66° 59' 


.143 " 


101° 17' 


120° 


.241 " 


60° 24' 


.126 " 


82° 52' 


150° 


.157 " 


55° 26' 


.043 " 


74° 51' 



HYDRAULIC MOTORS. 19 

According to this table the water is thrown backward, or 
in the direction opposite to the motion of the wheel for the 
outward How wheel, and for the inflow it is thrown forward for 
j'i less than 90 , and backward for y l greater than 120°. 

In the Tremont turbine a device was used for determining the 
direction of the water leaving the wheel, and for the best effi- 
.ciency, 79} per cent., the angle 6 was about 120°. Lowell 
Hydraulic Experiments, p. 33. 

The angle thus observed had a large range of values ranging 
from 5(V to 140° for efficiencies only two or three per cent, less 
than 79] per cent. 

13. Of the value of oo. 

So far as analysis indicates, the wheel may run at any speed ; 
but in order that the stream shall flow smoothly from the 
supply chamber into the bucket — thus practically maintaining 
the angles a and y Y — the relations in equations (5) and (6) must 
be maintained, or 

sin a jr oa . 

v. = — I , (oo) 

sin y x 

and this requires, that the velocity V shall be properly regu- 
lated, which can be done by regulating the head hi or the press- 
ure pi or both li x and p u as shown by equation (4). This 
however is not practical. In practice, the speed is regulated, 
and when the condition for maximum efficiency is established, 
the velocities V and v Y are found from equations (17) and (18). 
Since y 2i in practice, is small we have, for best effect, 

v 2 = Go'r 2i approximately, .... (37) 

and, adopting this value, a more simple expression may be 
found for the velocity of the wheel. For equation (19) gives 

v 2 = r 2 oo' 

: . (Approx.) (38) 



1/ 



cos asm y, /rA 2 ., sin 2 y. /r,\ 2 . 



sin(a + v 1 )\r , 2 / 2 'sin'- (a + y)\r 



20 HYDRAULIC MOTORS. 

If fj x = f < 2 = 0.10, r, -f-n == 1.40, a = 25°, y Y = 90°, the velo- 
city of the initial rim for outward flow will be 

" ri = whm = °- 929 v ^ 

The velocity due to the head would be 

v h = V2gff= 1.414 VgS; 
hence, the velocity of the initial rim should be about 

°- 928 ,g = 0.659 (39) 

1.414 VgR 

of the velocity due to the head. 

For an inflow wheel in which r* = 2r 2 2 , and the other dimen- 
sions, as given above, this becomes 

0.954 



1.414 



= 0.689 (40) 



of the velocity due to the head. 

The highest efficiency of the Tremont turbine, found experi- 
mentally, was 0.79375, and the corresponding velocity, 0.62645 
of the velocity due to the head, and for all velocities above and 
below this value the efficiency was less. Experiment showed 
that the velocity might be considerably larger or smaller than 
this amount without diminishing the efficiency very much. 

In the Tremont turbine it was found that if the velocity of 
the initial (or interior) rim was not less than 44 nor more than 
75 per cent, of that due to the fall, the efficiency was 75 per 
cent, or more. Exp,, p. 44. 

This wheel was allowed to run freely without any brake 
except its own friction, and the velocity of the initial rim was 
observed to be 1.335 V2gII, half of which is 



0.6675 V2gH, (41) 



HYDRAULIC MOTORS. 



21 



"which is riot far from the velocity giving maximum effect; 
that is fco say, when the gate isfuUy raised the coefficient of effect is 
a maximum when the wheel is moving with about half its maximum 
velocity" Exp., p. 37. 

M. Poncelet computed the theoretical useful effect of a 
certain turbine of which M. Morin had determined the value by 
experiment. The following are the results (Comptes Rendus, 
1838, Juillet) : 

TABLE V. 



Velocity o l initial 








rim or 


Number of turns of 

the wheel per 

minute. 


Ratio of useful to 


Means of values by 


/•jto' 


theoretical effect. 


experiment. 


J&JH 




0.0 


0.00 


0.000 




0/2 


33.80 


0.664 




0.4 


47.87 


0.773 


"oiioo" 


0.6 


58.61 


0.807 


0.705 


0.7 


62.81 


0.810 


0.700 


0.8 


67.67 


0.806 


0.675 


1.0 


75.76 


0.786 


0.610 


1.2 


82.88 


0.753 


0.490 


1.4 


89.52 


0.712 


0.360 


1.6 


95.70 


0.664 


0.280 


1.8 


101.51 


0.612 


0.203 


•2.0 


107.00 


0.546 


0.050 


3.72 


145.00 


0.000 






Poncelet states that he took no account of passive resist- 
ances, and hence his results should be larger than those of 
experiment as they are ; but here both theory and experiment 
give the maximum efficiency for a velocity of about 0.6 that 
due to the head, and the efficiency is but little less for velocities 
perceptibly greater and less than that for the best effect. For 
velocities considerably greater and less, theoretical results are 
much larger than those found by experiment, for reasons 
already given, chief of which is the fact that eddies are induced, 
and the effective angles of the mechanism changed to unknown 
values. 



22 HYDRAULIC MOTORS. 

14. Pressure in the wheel. 

Dropping the subscript 2 from v, r, p, in equation (9), the 
resulting value of p will give the pressure per unit at any point 
of the bucket providing that jj 2 be considered constant. Chang- 
ing r to p, equation (9) thus gives 

V>-(1 + V*)V 2 + Go'ip 2 -^ 2 ) 1 



P = 



2g + ^' * W 



v^v^ 


h ' sin ( 


..CO 


n 


From (4) and (5), 








Pi = #h + p a - 


d 1 + Mi 


sin 2 y 1 




sin 2 (a + 


n) 


These reduce equation (42) to 







To solve this requires a knowledge of the transverse sec- 
tions of the stream, for the velocity v will be inversely as the 
cross section. 

From equations (20) and (6) 

h_h sin a _ 

caVV . (44) 



- 1 [(1 + ^AV. + Wd + *) " Wy] ,^^ 

The back or concave side of the bucket will be subjected 
to a pressure which may be considered in two parts : one due 
to the deflection of the stream passing through it, the other to 
a pressure which is the same as that against the crown, and is 
uniform throughout the cross section of the bucket, due to the 
pressure of a part (or all) of the head in the supply chamber. 
It is the latter pressure which is given by the value of p in 
equation (45). The construction of the wheel being known, 
the pressure p may be found at any point of the wheel for any 
assumed practical velocity ; although, for reasons previously 



BYDItAULIC MOTORS. 23 

given, it will be of practical value only when running near the 
velocity for maximum efficiency. There are two cases: 

1. That in which the discharge is into free air; 

2. That in which the wheel is submerged. 

In the first case if the pressure is uniform, the case is 
called that of " free deviation " in which the entire pressure 
upon the forward side of the bucket is due to the deviation of 
the water from a right line, and will be considered further on. 

If equation (45) shows a continually decreasing pressure 
from the initial element to that of exit, or if the minimum 
pressure exceeds p a , the preceding analysis is applicable. 
But if it shows a point of minimum pressure less than ^ a , it 
will be in a condition of unstable equilibrium, in which the 
slightest inequality would cause air to rush in and restore the 
pressure to that of the atmosphere ; so that the pressure in 
the wheel and the flow would be changed. The point of mini- 
mum pressure may be found by plotting results found from 
equation (45), substituting values for p taken from measure- 
ments of the wheel, and h from computation. From the 
entrance of the wheel up to the point of minimum pressure 
the preceding analysis applies ; and the remainder of the wheel 
must be analyzed for " free deviation " and the two results added. 

In the second case the equations will apply, since air can- 
not enter, provided that p does not become negative, to realize 
which requires a tensile stress of the water. This is impos- 
sible and eddies would be formed ; and the effect of these on 
the velocity and pressure cannot be computed. Such a case 
cannot be analyzed. 

15. To find the pressure at the entrance to the bucket when 
running at best effect. In (45) let p — r ]} k — k { and p — p x . 
To simplify still more, let the wheel be frictionless, or Mi = 
fA 2 — 0, and find from equation (38) 

2 , 2 sin (a + v\) TT ,.„. 

r?oo - = ^ =— ^ all, (46) 

cos a sin y x 



24 HYDRAULIC MOTORS. 

also hi =■ H + h 2 , and (45) becomes 

Pi = dH+dh,+p a -^ dB ^p , ,5: . (47) 

1 x 2 cos or sin (or + Ki) v 7 

If the wheel is not submerged lu = 0, and let the pressure 
Pi equal that of the atmosphere, or p a , then 

0=1-3 Sl nr , 1 + , («) 

2 cos ex sin {a + Yi) 

If the wheel be submerged, let pi = Sh 2 + p a , and the equa- 
tion reduces to that of the preceding. 
Equation (48) gives 

tan 2^ = — tan y l9 
or, 2a = 180° -y x \ (49) 

for which value the pressure at the entrance to the wheel will 
equal that just outside. 

If, 2a > 180° - yi, (50) 

the pressure within will be less than that without ; but if 

2a < 180 = - yi, (51) 

the pressure within will exceed that without — a condition 
which is considered desirable. If frictional resistances be 
considered the value of rico from equation (38) will be less 
than that given by (46), and hence the last term of equation 
(47 ) will be less unless a be greater than the value given by 
equation (51 » ; hence with frictional resistances the terminal 
angle of the guide blade may exceed somewhat 90° — \y\ '■> 
therefore, if the value of ex be found for a frictionless wheel it 
will be a safe value when there is friction. If y = 90° and 
a = 90° - iYi = 45°, then (47) gives 

p l = dR + dhv +p a - 6H=z dhz + p a , . . (52i 



HYDRAULIC MOTORS. 25 

as it should.- [f y i = 90° and a = 30°, then 

or, ^> <M S +^>, -fO.SSrt//". 

Tlie angle <? should not be so large or y x so small as to pro- 
duce excessive pressure at the entrance to the wheel. 

Example. — Find the pressure per square inch at the 
entrance to the wheel when the head is 10 feet, the terminal 
angle of the guide is 30°, the initial angle of the bucket 
y 1 — 90 ; the wheel being one foot under the water in the tail 
race. 

16. Nu m be r of buckets. 

The analysis given above is true for a wheel with a single 
bucket, provided the supply is constantly open to the bucket 
and closed by the remainder of the wheel. But for practical 
considerations the wheel should be full of buckets, although 
the number cannot be determined by analysis. Successful 
wheels have been made in which the distance between the 
buckets was as small as 0.75 of an inch, and others as much as 
2.75 inches. LoweU Hyd. Exp., -p. 47. Turbines at the Centen- 
nial Exposition had buckets from 4:. inches to 9 inches from 
centre to centre. 

17. Ratio of radii. 

Theory does not limit the dimensions of the wheel. In 
practice, 

for outward flow, r 2 -f r x is from 1.25 to 1.50 I 
for inward flow, r 2 -f- 1\ is from 0.66 to 0.80 ) 



(54) 



It appears from Table II. that the inflow wheel has a 
higher efficiency than the outward flow wheel ^columns 6 and 
3), and these wheels have about the same outside and inside 
diameters. The inflow wheel also runs somewhat slower for 



26 HYDKAULIC MOTORS. 

best effect. The centrifugal force in the outward flow wheel 
tends to force the water outward faster than it would other- 
wise flow ; while in the inward flow wheel it has the contrary 
effect, acting as it does in opposition to the velocity in the 
buckets. 

It also appears that the efficiency of the outward flow wheel 
increases slightly as the width of the crown is less, and the 
velocity for maximum efficiency is slower ; while for the inflow 
wheel the efficiency slightly increases for increased width of 
crown and the velocity of the outer rim at the same time also 
increases. 

Let r\ — nr 2) y x — 90°, y 2 = 20°, Mi = fa — ; 

then for n = 0, 0.5, 0.8, 1, 1.4, co'r, = 2, 3, 

will oo'r. 2 = oo 

and E — 

18. Efficiency, E. 

The method of determining the theoretical value of E has 
already been given ; but to determine the actual value, resort 
must be had to experiments. These have been made in large 
numbers and the results published. By assuming the mini- 
mum values of the several losses, a maximum limit to the effi- 
ciency may be fixed. Thus, if the actual velocity be 0.97 of 
the theoretical, the energy lost will be (1 - 0.97 2 ) or 6 per cent. 
Friction along the buckets and bends . . . 5 " 

Energy lost by impact, say 2 " 

Energy lost in the escaping water .... 3 " 

Total .... 16 " 
Leaving 84 " 

available for work. This discards the friction of the mechan- 
ism and frictional losses along the guides, and if 2 per cent, 
be allowed for the latter, there will be left 82 per cent. It 



HYDRAULIC MOTORS. 27 

seems hardly possible for the effective efficiency to exceed 82 
per cent , and all claims of 90 or more per cent, for these 
motors should be at once discarded as being too improbable 
for serious consideration. A turbine yielding from 75 to 80 
per cent, is extremely good. The celebrated Tremont turbine 
gave 79j per cent. Lowell Exp., p. 33. Experiments with 
higher efficiencies have been reported. A Jonval turbine 
(parallel flow) was reported as yielding 0.75 to 0.90, but Morin 
suggested corrections reducing it to 0.63 to 0.71. (Weisbach, 
Mech. of Eng,, vol. ii., p. 501.) Weisbach gives the results of 
many experiments, in which the efficiency ranged from 50 to 
84 per cent. See pages 470, 500-507. See also Jour. Frank. 
Inst., 1843, for efficiencies from 64 to 75 per cent. Numerous 
experiments give E = 0.60 to 0.65. The efficiency, considering 
only the energy imparted to the wheel, will exceed by several 
per cent, the efficiency of the wheel, for the latter will include 
the friction of the support, and leakage at the joint between 
the sluice and wheel, which are not included in the former ; also 
as a plant the resistances and losses in the supply chamber are 
to be still further deducted. 

19. The Crowns. — The crowns may be plane annular discs, 
or conical, or curved. If the partitions forming the buckets be 
so thin that they may be discarded, the law of radial flow will 
be determined by the form of the crowns. If the crowns be 
plane, the radial flow (or radial component) will diminish as 
the distance from the axis increases — the buckets being full — 
for the annular space will be greater. 

20. Designing. 

The dimensions of a wheel must be determined for a definite 
velocity. Thus far it has been assumed that the angles or, y,, 
etc., are given, and the normal sections of the stream thus 
deduced. We will now assume that all the dimensions of 
the buckets are known, and the angle a and the section iT are 



28 HYDRAULIC MOTORS. 

to be determined. The velocities i\ and v 2 must now be found 
independently of a. From Fig. 1 we have 

V 2 — v 2 + oo 2 7\- — 2v 1 dv l cos ~/i ... (55) 

which combined with equations (4), (9), (10), vjc^ = v 2 k 2i as in 
(20), and H = h Y — /i 2 , will give 

hi _ (1 + /^i) fafaaor-^ cos y x 

V2 ~~ Jc 2 Vl ~ (1 + f.i 2 ) h 2 + \A x h} 






+ 



where 



"(1 + /<i) l\h 2 Gor x cos yi 
. (1 4- //,) K~ + i-ixkj . 

= Poo + ^/2gHQ + R 2 oo* (56) 

f jU/^faco Sj;i _ _ (56a) 
(1 + fa) fa + yWife 

= j— ^4 r- 2 (566) 

* (l + fj 2 ) fa 2 + ^ki K J 

v ' (?\~^:L^ +pt • • • (56c) 

^=h (57) 

Equation (11) gives the velocity of exit. Equation (14), or 
(130), gives 

u i r k 

E = SQR = gB[S ri2 ~ T ^ °° 2 + ^ 2 C ° S n ~ fa />l C ° S y ^> V ' l0D 

= ~[- S 2 go* + Tv 2 gj] (58) 

in which 

8 2 = r 2 2 -r t 2 (58a) 

fa 
T — r 2 cos y 2 — 77 n cos y\ (586) 

fC-2 



HYDRAULIC MOTORS, 29 

Substituting v 2 from equation (56) gives 



E= \ [(- S 2 + TP)g£ + T VZgHQc*? + 
<7^ 



= -U[- F 2 ^ + T V2gHQ<*> i fi 2 G**-\ . . . (59) 
in which 

F 2 = ^p _ ^2 # ( 59a) 

For is'a maximum, make -=— = 0, giving 

» ^ 2 V2glTQTTlh<S 2 = T{gHQ + B 2 go*); 



•••- ~^r v p-^ • • - (60) 

which is of the same form as equation (16). This value of go' 
substituted in equation (59) will give the maximum efficiency, or 

bl*. = (f» - vr i - r 2 /^ J. . . . (6i) 

For the frictionless wheel Q = 1. 

To find the terminable angle a of the guide blade that will 
enable the stream to flow smoothly, subject to the preceding 
conditions, Fig. 1 gives 

V cos ex = gdt^ — v x cos y l9 

which, combined with equation (55), gives 

GOTi — v Y cos y x , aox 

cos a = :. . • (62) 

yvi 2 + Gfr 2 — 2v 1 got 1 cos y x 

Eliminating v x by means of equations (57) and (56) gives 
cos ot in terms of the six constants r lf r 2 , yi, y^ h and & 2 > which 
are fixed and known from the dimensions of the wheel, and of 
the velocity go of the wheel. Since the wheel may run at dif- 



30 



HYDRAULIC MOTORS. 



ferent velocities the angle a must vary, and this will be done 
in practice by the piling of the water in the passages. Each 
turbine, however, should be designed to run at the speed giv- 
ing maximum efficiency, and its angles and dimensions should 
satisfy equations (60) and (62). 
From equation (9), 



2g^ = (l + M 2)v\-v 2 



a^ir 1 



+ 2? 



(63) 



in which, if v 2 and V\ be substituted from above, p x becomes 
known. 

Similarly, V from equation (55) becomes known, and finally, 
from (60), 



oor x — Vi cos y x 
cos a = — ^ r -\ 



(64) 



21. Path of the Water, — Let a A be the position of the bucket 

when the water enters at b. 
Q »> The bucket being drawn in 

position to a scale, divide it 
into any number of parts — 
equal or unequal — aa lf a x a 2> 
etc., and find the time re- 
quired for it to go from a to 
aj. The distance being small, 
assume that the velocity is 
uniform from a to a lt and 
equal to v, which will be 
given by equation (56) by 
dropping all the subscripts 2 
Fig. 7. and changing r to p, thus, 

_ (1 + fii) h x koori cos y x 




(1 + J*i) W + fix W 



t* 



gH+ <a 2 [/o»- (2 + #!)?,« ] 
(1 + yw a ) W + fix W 



kx* + 



' (1 + Hi) ki koori c os y 
. (1 + ft*) hx* + M 



08 yn * 

i w J • 



(64a) 



HYDRAULIC MOTORS. 31 

Then will the fcime / be 

t = (Uh (65) 

v 

During this time the rim has gone from a to b a distance 

ah = ?\ art (66) 

If the bucket bB be drawn through b, and the arc a v b x 
through a u their intersection b { will be the position of the 
particle at the end of the time t. In a similar manner, the 
successive points h,, b 3 , etc., may be found, through which a 
continuous curve may be drawn representing the path of the 
stream. 

The line tangent to the termination of the bucket, will 
indicate the direction of the water at entrance of the wheel, 
and if the water drives the wheel, the path should be entirely 
outside this line and convex toward it. 

22. Design a guide blade, outflow turbine. 

Assume the effective fall, H '= ft. 

Assume the required horse-power, . . . HP = 

Assume the exit angle, y 2 = 

Assume the entrance angle of bucket, . . y x = 90°. 

Fix the exit angle of the guide, Eqs. (32), (51), a = 30°. ? 

Assume efficiency, E = 0.65, or 0.70; 

and after the wheel is fully designed, re- 
compute this value and if necessary 
correct the dimensions. 

Eequired quantity of water per second 

without loss, Q z=z U ' -r- SH* 

Required quantity, Q -=- E = 

Assume }i x = 0.10, // 2 = 0.075. 

Velocity of the initial rim, Eq. (40), approx., oor x = 
(The corrected, final value will be found 
by Eq. (16) or (60). 



32 HYDKAl LIC MOTOKS. 

Let r 2 = 1.3/ 1 !, then velocity of outer rim, got 2 = 

The velocity into the bucket, Eq. (18;, . . v., ^ 

Initial section of buckets, Eq. (20), . . . & x = 

The inner circumference will be 2nr i . Let 
the walls of the buckets be T \ of the 
circumference, then the effective open- 
ings will be £f of the circumference, or 
i-f-nr^. Assume a depth, y, between 
the crowns. Try y = h\. Then will 
the initial cross section of all the 
buckets be {^nr? ; hence, \%7tr? — h L ; .\ r x = 

If the radius is not what is desired, it may 
be changed to some other value and 
the depth y computed. Then, . r 2 = 

The cross section at outer rim will be, if 

the crowns are planes, J s 5;rr 2 D sin y 2 = k 2 = 

The number of buckets will be assumed . = 

Having determined these elements, the final velocity v 2 in 
reference to the bucket may be confuted by equation (67), 
coi'i from (62), Ffrom (55), and ol from (64). 

If the turbine revolves in air, at least half the depth of the 
wheel is to be deducted from the head H. 

If the circular opening between the wheel and sluice be ^ 
of an inch, or j^j °f a f°°t, and the coefficient of discharge be 
0.7, the discharge will be 

^ 0.7 x 2*r, x /?fl =q, .... (67) 

Pi being determined from equation (63). The loss of work 
will be 

62.2g x II or 62.2? {H-%y) (68) 

The work lost by friction, if the radius of the axle be r B , the 



HYDRAULIC MOTORS. 



33 



weight of the loaded wheel, Jr 3 , and coefficient of friction // 3 , 

will be nearly 

2 

~ t<9W s r 3 M per sec (69) 

The work done by the water must be the effective U plus 
the work due to the losses. The work done by the water pass- 
ing through the wheel must be U plus that given by equation 
(69 1. Call this U x . Compute the work done by the water, and 
let it be U* ; then will the required depth be 



y-2 






(70) 



The efficiency may now be recomputed. 



SPECIAL WHEELS. 

23. Fourneyron Turbine. — All wheels having guide blades 
are of the Fourneyron type, although the wheels made by him 
were outward flow. The preceding 
analysis is a general solution of this 
turbine. 

24. Francis and Thomson's vortex 
wheds are inward flow wheels with 
guide blades. The preceding analysis 
is also applicable to these wheels. 

25. The Jonval Turbine is a parallel 
flow wheel with guide blades, to which 
the preceding analysis is applicable by making r t = r 2 . 

(For the details of these and many other forms, see 
Hydraulic Jlotors by Weisbach.) 




26. Rankine Wheel.— This is a wheel of the Fourneyron 
type, but Eankine having made certain modifications in its 
3 



34 HYDRAULIC MOTORS. 

assumed construction it is indicated by his own name. 
(Fig. 8.1 

It is an outflow wheel and the crowns are so made that 
the radial velocity of the water in passing through the wheel 
will be uniform. If x be the abscissa from the axis of the 
wheel to any point of the crown, and y the distance between 
the crowns at that point, v r the radial component of the 
velocity, then 

y ■ 2?rx • v r = Q, 

or,- yx — a constant, .... (71) 

which is the equation to an hyperbola referred to its asymp- 
totes. This determines -the form of the crowns. If the wheel 
were inward flow, the depth would be greatest at the inner 
rim. 

In this wheel the initial element of the bucket is radial, or 
y l = 90° ; and Rankine assumed that the velocity for best effect 
must be such that the water will quit the wheel radially, or 
6 — 90°. These conditions given, from Fig. 1 and equations 
(5), [6), (34), for & f motionless wheel, 

oovi = V cos a (72) 

oor 2 — v 2 cos y 2 (73) 

v x = w = Fsin a = got^ tan a — aor 2 tan y 2 = v 2 sin y 2 ; (74) 

.*. tan a — - tan y 2 (75) 

which determines the proper angle of the guide blade when 
the value of y 2 has been assigned. If y 2 = 15°, r 2 — \\r x , then 
a = m° ; and if y 2 = 20, r 2 = 1.3, then a = 24° nearly. But 
to be certain that the internal pressure exceeds the external, 
a should exceed these values. Equations (19), (73), and (75) 
give 



/ 



2gH 



which establishes the velocity of the initial rim of the wheel. 



HYDRAULIC MOTORS. 35 

The work in the frictionless wheel will be the theoretical 

work the water could do, less the energy in the water quitting 

the wheel, or 

W 

9 

w 

[Eq. (74)] =WH- J— n W tan 2 *. . . . (77) 

WH 

w*™ tS; < 78) 

The efficiency will be 

F =im = TTitetf^ (79) 

27. The following is to show that Eankine's assumption of 
velocity for best efficiency is not quite correct. Substituting 
y x — 90 : , /<! = 0, f.( 2 = 0, r x = nr 2 , in equation (16) gives 

^ = 9H f l - n 2 - V(T- n*) 2 - cosy, (1 - 2w«T ] 8Q 
1 - 2n 2 |_ V(l - n 2 ) 2 - cos 2 y 2 (1 - 2ri 2 ) J 

and these in equation (19) will give 

«.i™„--l/ rr , r i ~ n2+ A / (l-^) 2 -COS 2 r 2 (l-27l 2 n 

v 2 cos r2 -y ^^cos^/J ' 9 ,-, r 

r L V (1 — w) " — cos- y 2 (1 — 2n 2 ) J 

= cor 2 [1 - n 2 + V{l-n 2 ) 2 - cos 2 r 2 (1 - 2rc 2 )]. . (81) 

This does not give 

v 2 cos ;/ 2 = oor 2 

as in equation (73), except when n 2 = J (or 2r 2 x = r 2 2 ) ; and hence 
the direction of the water at exit will not be radial, as Eankine 
assumed, except for this case. In practice, outflow wheels are 
constructed almost exactly with this proportion, n 2 — J, and 
hence the analysis from equations (72) to (79) is sufficiently 



36 



HYDRAULIC MOTORS. 



exact for the frictionless outflow wheel ; and, as seen above, 
the hypothesis greatly simplifies the analysis. 

The condition for best efficiency of the frictionless wheel 
requires that the velocity of leaving the wheel should be a 
mininiuni ; and this may be realized, in some cases, when its 
direction is oblique to the radius. 

Thus, let A be radial when AB is the velocity relative to 
the bucket, and B C the velocity of the rim ; then it may be, in 
some cases, that when AD is the relative velocity of exit, AE, 
the velocity of exit relative to the earth, will be less than A C 9 
as shown in Fig. 9. 





Fm. 9. 



Fig. 10. 



28. The path of the ivater is easily constructed for this wheel. 
Since the radial velocity is uniform, the time of flowing through 

the wheel will be 

r 2 — n 



t = 



Vi 



(82) 



during which time the initial rim AB, Fig. 10, will have travelled 

aB = oor x t feet (83) 

Divide r 2 — r x into equal parts by concentric arcs, and the 
space aB into the same number of equal parts, and through 
the points of division a, &, c, d, trace the buckets ; then will aD, 
drawn through the proper intersections of the arcs and buckets, 
be the required path. 



HYDRAULIC MOTORS. 37 

9. Analyze a Rankine turbine, having given : H = 12 feet, 

y% - 15 : , /', = 2 feet, r 1 , = % ; \. Depth of outer rim, 6 inches. 

Find Radius of outer rim, r 2 — 

Angular velocity, go — 

Velocity of initial rim, .... ?\oo — 

Velocity of outer rim, r 2 oo = 

Angle of guide plates, a = 

Velocity from the supply chamber, . F = 
Initial velocity in bucket, . . . . v x = 
Terminal velocity in bucket, . . . . v 2 = 
Velocity of quitting water, . . . . w = 

Depth of inner rim, y x = 

The horse-power, HP = 

The efficiency, E — 

If the partitions for the buckets occupy gV of the wheel, and 

the losses due to frictional resistances in the wheel and friction 

of the wheel be 20 per cent., what will be 

The horse-power, HP = 

The efficiency, E = 

Find the pressure at the inner rim, . . . pi — 

Find the path of the water. 

29. Velocity of a particle along a tube rotating about an axis 
perpendicular to its j)lane. 

This problem has already been solved in establishing the 
general equations of turbines, and the following is given to 
present it from another point of view. 

If the particle at A, whose mass is m, be confined while the 
tube rotates about 0, Fig. 11, with the angular velocity go, the 
centrifugal force would be 

f=mco 2 p (84) 

If 6 be the angle between the radius vector prolonged and 
the normal upon the tangent, the component in the direction 
of the tangent to the tube will be 



38 



HYDKAULIC MOTORS. 

mcJp sin 0, 



and when the particle is free to move, this component will be 
effective for producing motion, and if the pressures at the 
opposite ends of the element are not equal, but differ by an 
amount dp, we have the equation 



But 



d?s 
m-rp z = mo6 2 p sin 6 — dp. 



(85) 





Fig. 12. 



Fig. 11. 



ds sin 8 = dp, 



which combined with the preceding equation gives 



dt 



2 j ! dp J 

= G0 2 pdp : - a dp. 

m sin x 



. . (86) 



But -J?— n = ds, and if 6 be the weight of unity of volume, 

srn# 

then m = -ds, and the last term becomes | dp. The integral 

if 

will be 



df} = 



a _ «.* = at (r* — «*.\ — O-P? & 



w, 8 = ca 2 (r 2 2 



2sr- 



(87) 



HYDRAULIC MOTORS. 39 

If the friction be /' 2 'Y, the equation becomes 

(1 + //.,) vf = v? + ^ {ri - r, 2 ) - 2^ ^T^ 1 , . (88) 

which is the same as equation (9). 
If p 2 = p { , and ;/ 2 — 0, then 

V.? = !>!« + &9 2 (r 2 2 - 7^) (89) 

This gives the velocity relative to the tube whether it 
revolves to the right or left, and whatever be its curvature. If 
it revolves to .the left, the resultant velocity will be AD, Fig. 
12 ; if to the right, it will be A C. If y 2 be measured from the 
arc backward of the motion, or y 2 = BAF for rotation to the 
left, and y 2 = EAB for motion to the right ; then 

AD 1 = id 2 = v-2 + oo'W — 2v 2 . oor 2 cos y 2 . . . (90) 

A C 2 = iir - v} + oo 2 r 2 2 + 2v 2 . oor 2 cos y 2 . . . (91) 

In the latter case the quitting velocity will exceed the ter- 
minal velocity in the tube, and therefore increased velocity will 
have been imparted to the water — a condition requiring that 
energy be imparted to the wheel from an external source. In 
the former case the wheel is a motor, in the latter it is a re- 
ceiver or transmitter of power ; in the former the water drives 
the wheel, in the latter the wheel drives the water and virtually 
becomes a centrifugal pump. 

If the water issues tangentially to the path described by the 
orifice, then y 2 — 0, and 

w — v 2 T oor 2 , (92) 

the upper sign belonging to the motor, and the lower to the 
pump. 

Exercise. — If r x — 1 ft. r 2 — 5 ft jj 2 = 0.1, v x — 5 ft. per sec- 
ond, and the bucket rotates about a vertical axis 30 times per 



40 HYDRAULIC MOTORS. 

minute, and discharges the water directly backward, making 
y 2 = 0, required the terminal velocity along the tube and the 
velocity of discharge relatively to the earth 

30. Wheel of Free Deviation. — In this wheel the water in the 
buckets has a free surface, or, in other words, is subjected only 
to the pressure of the atmosphere. For this case 

Pi — Pi = p a ) h — H, and k 2 = 0, 

and equation (4) gives 

(1 + yuO F 2 = 2(//7, ..... (93) 

which will be the velocity of discharge from the supply 
chamber into the wheel ; it is the velocity due to the head in 
the supply chamber when frictional resistance is included. 
The triangle ABC, Fig. 1, gives 

v 2 = V 2 + oo 2 r 2 - 2 Ywr, cos a, ... (94) 

which substituted in equation (88) gives 

(1 + ;/ 2 ) v 2 = V 2 + go 2 n 2 - 2 V. av, cos a, . (95) 

and this in equation (90) gives w 2 , and equation (12) will give 
the required work. Equation (14) will give the velocity for 
best effect. But this involves a long analysis, and the follow- 
ing approximate solution is sufficiently accurate. 

If y 2 be small, and the wheel be run for best effect, that is, 
so as to make the velocity to very small, and considering w = 0, 
equation (92) makes 

v 2 = GDV 2 nearly. 

Using this value as if it were the exact one, also neglecting 
friction, (95) gives 

2 V. gdT] cos a — V 2 — 2gH t 
or 2 awi cos (x = \/2gJI; 

,\ gdt 1 = — '^— ■ (approx.) .... (96) 

COS a ^ r A ' 



HYDRAULIC MOTORS. 



41 



which gives the proper velocity of the initial rim ; and for the 
terminal rim 



cor, 



_ Vijjr//. 



cos a r x 
Number of revolutions per minute 



.Y=30". 



(97) 



(98) 



To find the velocity at any point of the bucket relative to the 
bucket, drop the subscript 2 from equation (95) giving 

■ (99) 



(1 -f yw 2 ) v- — 2gII + go 2 r 2 2 — 2 V^gH. oor l cos 
A 





Fig. 14. 



Fig. 13. 
From Fig. 1 or equations (4) and (5) find 



V2gR . 
sin i/i = — sin a. 



In the frictionless wheel, the work done will be 

U = i3f(V 2 -w 2 ) 9 . . 
and the efliciency will be 

U 



E = 



8QH 



(100) 

(101) 

(102) 



31. To find the form of the free surface, let the bucket be very 
narrow, so that a normal to one of the curves will be approx- 
imately normal to the other. Divide one side of the bucket 



42 HYDRAULIC MOTORS. 

into any convenient number of parts, as ac, ce, etc., and erect 
normals to the arc, as a ] >, ed, etc. Lay off these arcs on a right 
line. Compute the velocity at any point, as d, Fig. 13, by for- 
mula (99). Let x be the required depth at d, then because the 
velocity into the section equals q, the volume passing through 
one of the buckets per second, we have 

x . dc . v = q ; 

and similarly for all other sections. If only relative heights 
are to be found, the quantity q need not be found, for if y be 
the height at b, Fig. 14, then 

y.la.Vy = q; 

ba.v l , . 

- a!= ^r^ •••••• um) 

and by assuming any arbitrary value for y the relative value of 
x becomes known. Similarly, the relative heights at all other 
sections may be found. 

32. To find the path of the fluid in reference to the earth, pro- 
ceed as in Article 21 of the discussion of the general case. 

33. Exercise. — Design a 30 horse-power inflow turbine of 
free deviation, given an effective head of 16 feet. 

Assume the depth of gate opening to be 4 inches (^ foot), 
and after the computation has been completed if it does not 
give 30 horse-power the depth may be changed by proportion. 
Let the radius of the outer or initial rim be 1 ft. ; of the inner 
rim, | of a foot ; terminal angle of the bucket, y 2 = 15° ; termi- 
nal angle of the guide, a = 30°, j.i v = 0.10 — yu 2 . 

Then, velocity of exit from supply chamber, 

Eq. (93), V= 

Velocity of outer rim, Eq. (96), .... oo^ = 
Velocity of inner rim, Eq. (97), .... oor 2 = 



HYDRAULIC MOTORS. 



43 



Number of turns per minute, = 

Initial angle of bucket, Eq. (100), . . . y x = 

Initial velocity in bucket, Eq. (94), ... v x = 

Terminal velocity in bucket, Eq. (95), . . v 2 == 

Velocity of exit, Eq. (90), w = 

Direction of outflow, Eq. (35), .... = 
Coefficient of discharge 0.60, volume of 

water, Q = 

Weight of water (S = 62.4), 6Q = 

Work per second, Eq. (101), U = 

Horse-power, HP — 

Efficiency, . . E ' = 

If 90 per cent, of U is effective work, and if this does not 
give 30 horse-power, then the depth of the wheel should be 

d — TTTT77 ^ i ncnes - 

Find the profile of the stream in the buckets. 

34. The following is taken from the report of the Commis- 
sioners of the Centennial Exposition, 1876, on Turbines, Group 
XX. The tests were for two minutes each. The revolutions 
and horse-powers here given are those corresponding to the 
best efficiencies : 



Diameter 
of wheel. 
Inches. 


Head in sup- 
ply chamber. 
Feet. 


Revolutions 
per minute. 


Horse- 
power. 


Efficiency, 
per cent. 


No. of 
Buckets. 


Kind of wheel. 


30 


31 


255 


95 


85.0 


10 


Inflow. 


24 


31 


302 


67 


77.0 


14 


Parallel. 


24 


30£ 


310 


64 


74.5 


13 




27 


30 


291 


76.8 


80.3 


16 




30 


30 


257 


74 


75.5 


18 




25 


31 


288 


46 


82.0 


12 


Parallel. 


30 


29.2 


258 


80.5 


78.7 


13 


In and down 


25 


30 


279 


62.5 


83.7 


15 


In and down. 


27 


30.4 


246 


53.2 


73.6 


14 


Parallel. 


36 


29.6. 


197 


66.2 


83.8 


26 


Parallel. 



44 



HYDRAULIC MOTOES. 



These tests were by no means exhaustive. It is not known 
that they were run for best effect. The distance from centre 
to centre of buckets varied from 4.3 inches to 9.5, and at these 
extreme values the efficiencies were about the same. The 
number of gate openings was less than the number of buckets. 



TUEBIKES WITHOUT GUIDES. 

35. Barker's Mill. — As ordinarily constructed, this motor 
has two hollow arms connected with a central supply chamber, 

with orifices near their 



outer ends and on oppo- 
site sides of the arms. 
There are no guide 
plates The supply 
chamber rotates with 
the arms. The arms 
may be cylindrical, con- 
ical, or other convenient 
shape. 

Since the water issues 
perpendicularly to the 
arms y- 2 = ; and since the initial elements of the arms are 
radial, y± = 90°, and as the water must flow radially into the 
arms, ex — 90°. The inner radius is necessarily small and may 
be considered zero. Hence, making 

Y2 = 0, ri = 90°, a = 90°, n = 0, 




in equation (14) gives 

_ U _ GOTz 

*-*QH. gH 

Equation (19) gives 



aor 2 + 



\/ML 



+ afr? 



1 + \x 2 






. (105) 



(106) 



HYDRAULIC MOTORS. 45 

hence, the efficiency reduces to, for the frictionless wheel, 

E=^- (107) 

V 2 + &>T 2 v 

This has no algebraic maximum, but approaches unity as 
the velocity increases indefinitely. Practically it has been 
found that the best effect is produced when the velocity of the 
orifices is about that due to the head, or 

gdt 2 - V2gH; (108) 

for which value the efficiency will be, if yu 2 = 0.10 

E= 2 [~- 1 + V n] = 070 (109) 

If k 2 be the area of the effective section of the orifice, then 

Q = k 2 v 2 (110) 

The pressure on the back side the arms opposite the 
orifices and useful in driving the mill, will be 

P 1 = Mv 2 = ^-v 2 (Ill) 

Of this pressure there will be required 

P 2 = M.Gor 2 = 6 Q.Gor 2 , (112) 

to impart to the water the rotary velocity oor 2 which it has 
when it reaches the orifice. The effective pressure will be 
P x - P 2i and the work done per second will be this pressure 
into the distance it traverses per second, or 

U=cor 2 [P l -P 2 l 

which reduces to the value found from equations (105) and (106). 

36. Exercise. — Let the supply chamber be square, and from 
two of its opposite sides let pyramidal arms project. Let 



46 HYDRAULIC MOTORS. 

11=10 feet, orifices each 2 square inches, vertical section of 
arms through the orifice each 4 square inches, section of the 
arms where they join the supply chamber each 8 square inches, 
horizontal section of the supply chamber 36 square inches, 
r 2 = 36 inches, velocity of the orifice oor 2 = \ / "2gH i coefficient 
of discharge 0.64, and jj 2 = 0.10. 

Be quired : 

Velocity of discharge relative to the orifice, v 2 = 

Velocity of discharge relative to the earth, w = 

Velocity at entrance to the arms, ... v x = 

Velocity in the supply chamber, ... = 

The volume of water discharged, ... Q = 
The weight of water discharged, . . . d Q = 

The work per second, . U= 

The horse-power, HP = 

The efficiency, . E = 

The pressure on arm opposite orifice at A 

per square inch, pi = 

The pressure at base of the arms at C, . p = 
The equation to the path of the fluid. 

37. Scottish and Whitelaiv Turbines. — These wheels have no 
guide plates, and differ from Barker's mill chiefly in having 
curved arms. The analysis is precisely the same as for the 
Barker's mill. The only practical difference consists in pro- 
viding a curved path for the water, instead of compelling the 
water to seek its path, forming eddies, etc. 

38. Jet Propeller. — We first show how this problem may be 
solved by the preceding equations, and afterwards make an 
independent solution. Let a narrow vessel, Fig. 16, be carried 
by an arm E about a shaft BA. Let water, by any suitable 
device, be dropped into the vessel, the horizontal velocity of 
the water being the same as that of the vessel. At F, the 
lower end of this chamber, let there be an orifice from which 
water may issue horizontally. The water may then be con- 



HYDRAULIC MOTORS. 



47 



sidered as entering the vessel or bucket without velocity, and 
passing downward finally curve towards, and issue from, the 
orifice. It thus becomes a parallel flow wheel without guides, 
and we have, for the frictionless wheel, 

>'i = r,, Y\ = 90 °> n = °> Mi = fh = 0, II — 0, 
Pz = Pi = Pa, <h = 

in equation (8) ; hence, the velocity of exit relative to the ori- 
fice will be 



v-2 — 2gz 2 



(113) 



I 





Fig. 17. 



Fig. 16. 

where z 2 equals the head in the supply chamber. Under these 
conditions the velocity of discharge will be independent of the 
velocity of rotation, if the rotation be uniform. 

Equation (11) gives for the velocity of discharge relative to 
the earth 

(1U) 



w 



Vo — con 



Equations (19) and (14) give 



TT S Q 

Ui = — - v 2 • t 2 go. 



(115) 



48 HYDRAULIC MOTORS. 

This equation may be factored thus, — — is the mass of 

liquid flowing out per second; represent by M; Mv 2 is the 
momentum of the outflowing liquid per second. Mv 2 r 2 is the 
moment of the momentum, and, finally, 

Mv 2 - t 2 gd is the moment of the momentum into the angular 
velocity, and equals the work done. 

Let v = Gor 2 = the velocity of the vessel ; then from (114) 
and (115) 

w = v 2 — v, . . . . . . . (116) 

U x = Mv 2 v; (117) 

which equations are true whether the motion be circular, 
linear, or in any other path. 

In practice, the velocity of the jet is produced by the press- 
ure exerted by a pump, in which case z 2 in equation (113) 
would be replaced by a virtual head, Fig. 17, equivalent to 
s 2 ; or 

« 2 2 =2 ? f (118) 

Also the vessel, instead of having water supplied to it at 
the velocity of the vessel, picks it up from a body of water 
considered at rest ; thus imparting to the water the momentum 
Mv, requiring the work per second 

U 2 = Mv 2 (119) 

Hence the effective work done by a jet propeller picking 
tip the water from a state of rest will be 

U= U x - U 2 = M(v 2 -v)v. .. . . . (120) 

The energy exerted by the pump will be that producing the 
velocity of water relative to the earth, or \ M(v 2 — v) 2 , plus 



HYDRAULIC MOTORS. 



49 



that doing the work of driving the vessel ; hence, the energy 
expended will be 

lM(v 2 -vf+ U; 

and the efficiency will be 



E= 



U 



2v 



U + ^M{y 2 — v) 2 v 2 + v 



(121) 



This has no algebraic maximum, but approaches unity as 
v f the velocity of the vessel, in reference to the earth, ap- 
proaches v 2 in value, the velocity of the jet in the opposite 
direction relative to the orifice. 

If v 2 = v, the efficiency will be perfect as shown by (120), but 
no work will be done as shown by (119). This would be the 
case of a vessel drawn by an external agency, or even floating 
along a stream ; for the water backward relative to the vessel 
would equal the forward velocity of the vessel. 

The mass of the jet per second will vary as the section of 
the orifice and velocity of the jet ; and if lc be the section of the 
jet, then 



U=-kv 2 (v 2 



v)v; 



(122) 



hence the same work may be done by enlarging the section k, 
and properly diminishing the velocity v 2 of the jet ; but as v 2 
is diminished, the efficiency is increased, as shown by equation 
(121). 

If v = 10 feet per second (about 6.8 miles per hour), we find 



r 2 


U 


Jt 


E 


10 


9 
750 &" 


00 


1.00 


15 


120.0 


0.80 


20 


2000 ifc" 


45.0 


0.67 


30 


6000 k" 


15.0 


0.50 


40 


12000 k " 


7.5 


0.40 


100 


90000 jfc" 


1.0 


0.16 



50 HYDRAULIC MOTORS. 

The sections k here given are for equal works, TJ. If the 
Telocity of exit be constant, then will the work increase 
directly as the area of the section while the efficiency remains 
the same. These are without frictional resistances. 

The pressure against the side of the vessel opposite the 
orifice due to the reaction of the water will be found from 
equation (117) by dividing the work done by the space over 
which the work is done, or 

P, = A = Mv K (123) 

v 

which is the momentum of the jet per second relative to the orifice. 
To impart to the water taken up the uniform velocity v 
would require the constant pressure 

P 2 = Mv ; 

hence the resultant pressure producing work would be 

P = P l -T 2 = M (n 2 - v), . . . . (124) 

and the resultant work would be 

T 2 = M (v 2 - vjv, (125) 

as before found in equation (120). 

The speed of a jet propeller depends upon the form of the 
vessel and the nature of the fluid ; but the pressures due to 
the action of the jet will be the same whether it issue into a 
vacuum, or into air or water, or a more viscous fluid. If a block 
be placed before the jet so close to the vessel as to obstruct the 
flow of water as a jet, the conditions will be changed, and the 
forward pressure will then be due partly to the direct pressure 
exerted by the pumps. If a piston, having a long piston-rod 
projecting against a firm body outside the vessel, be forced 
backward, the forward pressure, effective in driving the vessel, 
would be that exerted by the pumps less the frictional resist- 
ances. 



HYDRAULIC MOTORS. 51 

Tke efficiency of the jet propeller as a motor is compar- 
atively small in practice. This is due to the great loss of 
energy in the jet. The entire energy in the jet is lost. If the 
vessel be anchored, and the velocity of the jet be v 2 , the press- 
ure will be 

Pi = Mv 2 , 
the work will be 

and the 

energy lost = \Mv 2 , 

If the speed v of the vessel is small, then 
P l = Mv 2 , 

U = Mv 2 v, nearly, 

energy lost — %Mv 2 2 , nearly, 

and the energy lost will generally exceed considerably the use- 
ful work. 



52 



HYDEAULIC MOTORS. 



39. The difference of the moment of the momentum of the water 
on entering and leaving the wheel, equals the moment exerted by it on 
the ivheel. (Proof for the Motionless turbine by J. Lester 
Woodbridge, graduate of Stevens Institute, 1886.) 

Consider the effect of water passing along a smooth, curved 
horizontal tube rotating about a vertical axis. Conceive the 
water to be divided into an infinite number of filaments by 




vanes, similar to those of the wheel, but subjected to the con- 
dition that, at each point, their width, id, Fig, 18, measured on 
the arc, whose centre is 0, shall subtend at the centre a 
constant angle dd. Conceive each filament to be divided into 
small prisms, whose bases are represented by the shaded areas 
a'b'c'd', d'c'c'd" and abed, by vertical planes normal to the vanes 
making the divisions ae, ef, intercepted on the radius by circles 
passing through the consecutive vertices on the same vane, a, 
d\ d", etc., equal. 






HYDRAULIC MOTORS. 53 

Let p = the radius vector ; 

r = the height of an elementary prism ; 
then, dp — ote, ef, etc. ; 

pdOdp = abed, etc., = area of the base of an infinitesimal 

prism ; 
xpdOdp = volume of an infinitesimal prism ; 

d6dp= m — the mass of prism, 6 being its density 
or mass of unit volume ; 
y — san = angle between the normal to the vane at any 
point, and the radius Oa prolonged through that 
point ; 
X — velocity of a particle along the vane at p, which is 
assumed to be the same in all the vanes at the same 
distance from the centre ; 
go = the uniform angular velocity of the wheel, and 
p — the pressure of the water at the point p due to a 
head, but not due to deflection. 
Let p be the independent variable, and dt the time required 
for the element abed' to move its own length, dp, and ad the 
distance passed through by this element circumferentially in 
the same time, dt, then 

at do 



v sm y 
and, 

ad — oopdt = Gjp-j-dp. 

The mass m will have two motions : one along the vane, 
the other with the wheel perpendicular to the radius. By 
changing its position successively in each of these directions, 
both its velocity with the wheel and its velocity along the vane 
may suffer changes both in amount and direction, as follows : 
(I.) By moving from a to a, Fig. 18, in the arc of a circle — 
(1.) cop may be increased or diminished ; 
(2.) oo p may be changed in direction ; 



54 HYDRAULIC MOTORS 

(3.) v may be increased or diminished; 
(4.) v may be changed in direction. 
(II.) By moving from a' to d' along the vane — 
(5.) od p may be increased or diminished ; 
(6.) cop may be changed in direction ; 
(7.) v may be increased or diminished ; 
(8.) v may be changed in direction. 
These changes give rise to corresponding reactions, as 
follows : 

(No. 1.) Since the element is to move from a to a in the 
arc of a circle, oop will be constant, and hence the reaction 
= 0. 

(No. 2.) By moving from a to a', the velocity oop is changed 
in direction from ok to a'k' in the time dt. The momentum 
is moop, and the rate of angular change is 

kak! __ osdt _ 
~W~~aT~ m ' 

and hence the force upon the element producing motion in the 
arc of a circle will be radially inward and the reaction will 
be mGD 2 p radially outward. This is generally called the centrif- 
ugal force, as designated by most writers. Resolving into 
two components, we have 

moo 2 p sin y along the vane, 
moo 2 p cos y normal to the va,ne. 

(No. 3.) According to the conditions imposed, this value of 
v is the same at a 1 as at a, hence, for this case, the reaction 
will be zero. 

(No. 4.) In moving from a to a the velocity along the vane, 
v, is changed in direction from at to at' at the rate go as in 
No. 2. 

The momentum is mv, and the force will be mvao, which 
acts in the direction bn. Since the particle will be driven by 



HYDRAULIC MOTORS. 55 

the vane XY, and the reaction will be in the direction nb\ 
which being resolved, gives 

< > along the vane, 
— mvGj normal to the vane. 

(No. 5.) In passing from a' to d' ; at d' the circular velocity 
will be greater than at a' by the amount 

at dp, 
and the acceleration will be 

dp 

requiring a force moo tt tangentially to the wheel in the direc- 
tion of motion, the reaction of which will be 

dp 

but backwards, and its components will be 

moo -jr cos y along the vane, 

— moo — sin y normal to the vane. 
dt 

(No. 6.) In passing from a to d', oop will be changed in 
direction by the angle between Tea and k"d', or 

, n ,, _a'g _dp cot y 

p p 

and the rate of angular change will be 

cot y dp 
~~/T 'dt' 



56 HYDRAULIC MOTORS. 

and the momentum being 

moop, 



the reaction will be 



dp 
moo cot y -T7, 



which acts radially inward and its components are 

— moo cos y -T. along the vane, 

— moo cot y cos y -j- normal to the vane. 

dt 

(No. 7.) By moving from a to d\ v will be increased by an 
amount 

dv , 

a P df> > 

in the time dt, and the reaction will be 

dv do 
dp' dtf 

which will be outward along the vane, and the reaction will be 
directly backward along the vane, and hence is 

dv dp , ,, 
— m -=- . ^r; alone the vane, 
dp dt 5 

normal to the vane. 

(No. 8.) In passing from a to d\ v is changed in direction 
by two amounts : the angle y changes an amount 

<Ky"-y>)=-p P . 



HYDRAULIC MOTORS. 57 

This is negative, for a differential is the limiting value of 
the second state minus the first, and the first is here larger. 

But this is not the total change, since y" is measured from 
a radius making an angle 

dp cot y 

with Oa as in No. 6; hence the total change will be the sum 
of these, and the rate of change will be the sum divided by dt, 
which result, multiplied by the momentum mv, will give the 
reaction, which will be normal and in the direction b'ri or 



mv 



along the vane, 

"cot y dp dy dp~\ , , , , 

. -T- — ~ . -r, , normal to the vane. 

. p dt dp dtj 



This completes the reactions. Next consider the pressure 
in the wheel. The intensity of the pressure on the two sides 
ab and cd differs by an amount 

The area of the face is de x x = xpdB sin y, and the force due 
to the difference of pressures will be 

xpdO sin v ~- dp. 

' dp 

If dp is positive, which will be the case when the pressure 
on dc exceeds that on ab, the force acts backwards, and the 
preceding expression will be minus along the vane. In regard 
to the pressure normal to the vane, if a uniform pressure p 
existed from one end of the vane V TT to the other, the resultant 
effect would be zero, since the pressure in one direction on V W 
would equal the opposite pressure on XY. If, however, in 
passing from d to #, the pressure increases by an amount — dp, 



58 



HYDRAULIC MOTORS. 



since Va is longer than Xb, the pressure on Va will exceed that 
on Xb by an amount 

— dp . x x ah = — dp.x. pdd cos y — — xp cos ydO -±- dp. 
Collecting these several reactions, we have 



NORMAL TO THE VANE. 

(1.) 

(2.) + mou 2 u cos y. 

(4.) — juojv. 



(5.) — moo sin y 



dp 
TV 



(6.) — moo cot y cos y 



dp 
dt' 



(7.) 

fe% N fcot y dp dy dpi 
(8.) + mv . -jr- 4- . 4r I 

(9.) — #p cos y j- dpdd. 



ALONG THE VANE. 



-f moo-p sin y. 

0. 

dp 

+ racy COS y-rr. 

dp 
— mc^cos y-5T. 



W2 



^ ' dp* 



0. 



xp sin y -y- dpdd 



The sum of the quantities in the second column, neglecting 
friction, will be zero ; hence 

dp dv dp j Jn /ACkri 

moo-p sin y — m -^- ■ -* xp sin y -~- dpdu = 0. . (123) 



Substituting 



-j = v sin y, and xpdOdp = -^ 



and dividing by m sin y, we have 



oo 2 pdp — —dp = vdv. 



(124) 



HYDRAULIC MOTORS. 59 

Integrating, 

r n limit T- -| limit 

IWp'-P =\iv>\ .... (125) 

L- °-l limit L. J limit 

The sum of the quantities in the first column gives the 
pressure normal to the vane, which, multiplied by p sin y, gives 
the moment. This done, we have 



dr2T = mv sin p 



ooy (— go cos y — 2 J — pv sin y 

cos y dp 
+ v cos y — p sr- -/- 



dy 



Putting mv sin y = —^-dpdft, where Q is the quantity of water 

flowing through the wheel per second, and integrating in refer- 
ence to 6 between and 27T, we have 



dM=dQ\ copl -go cos y—2j — pvsmy-J- + vco&y — p ^ y ~ 
Multiplying (124) by 



dp 



P 

— cos y, 



we have 



&rp 2 7 p cos y dp 7 dv , 
cos yap r — -f dp = p cos y ^-dp 

v r vd dp r dp 



which substituted above gives 

a\3/=d() —2Gopdp + pcosy-j-dp + vcosydp—pvsmy-f-dp (126) 

the integral of which is 

M = SQ [— Gjp 2 + pv cos y] 

= -tfgp[«p-t>eoBrti88:. • • (127) 



60 HYDRAULIC MOTORS. 

But go p — v cos y is the circumferential velocity in space of 
the water at any point, and $ Q p [go p — v cos 7] is the moment 
of the momentum; hence, integrating between limits for inner 
and outer rims, the moment exerted by the water on the wheel 
equals the difference in its moment of momentum on entering 
and leaving wheel. 

Let the values of the variables at the entrance of the wheel 
be p„ y„ v^jpu and at exit p 2 , y„ v„ p. 2 . 

Equations (125) and (127) become 

i co> ( P : - p:) ._ £_p& = 1 « _ <). . . (128) 

M = S Q [go {p* - p 2 2 ) - Pi Vi cos y x + p 2 v 2 cos yj: (129) 

U = Mao = 6 Q go [go (p, 2 — p 2 2 ) — p l v 1 cos y 1 + p 2 v 2 cos y 2J (130) 

which is essentially the same as equation (58). It, however, in- 
volves the velocity of entrance, v iy and of exit, v 2 . The former 
may be found by equation (6), when a is known, or assumed 
when a is to be determined, and v 2 may be found by (19) or (56). 
This principle does not appear to be of great value in the solution 
of the general problem, but may be of much service in certain 
special cases. Thus, in the Barker Mill, page 44, the moment of 
the momentum of the water entering the wheel will be zero, but 
of exit will be 

M V, 

where V is the velocity of exit, relative to the earth, perpendicular 
to the arm, and the moment will be 

M Vr 2 ; 
.-. U = M V • r 2 go 

= Ps, (131) 

where P is the pressure on the arm opposite the orifice, and s 
the space passed over by the orifice in a second of time. 



HYDRAULIC MOTORS. 61 

But 



M 



V = >\ : _ r , co : 
6 Q 6 \ y 8 

(1 + M) v, = 2 g H + or r 2 2 ; 

# A' v. 
.-. U = LJ (r, - /<, c») /•„ <* (132) 

9 



6Q / t /ZgH+a?r? 






as in equation (105). 



40. Again, if the water quits the wheel radially, then the 
moment of the momentum of the quitting water will be zero, and 

U = 31 V )\ cos a ■ go. 
But 

V cos a = Y t , 

the tangential component of the velocity, or velocity of whirl ; 

.-. U = 31 r t i\ go. (133) 

41. In the frictionless Rankine wheel the velocity of whirl 
equals the velocity of the initial rim of the wheel. 

.-. V t = r x co; 
.-. U = Mr? oo\ (134) 

The work will also equal the potential energy of the water, 
W H = C) Q H, less the kinetic energy of the quitting water, 
\ 31 w* (less the energy lost in resistances, ju v*, which in this 
case we neglect) ; 

.-. U = 6 Q H - \3Iw\ 

and since the water is assumed to quit radially 

w = i\ go ■ tan y 2 = r x co tan a. t (135) 



62 HYDRAULIC MOTORS. 

The three preceding equations give 



2 -f- tan 2 a 



r, a» ={/ - 



as in equation (76). 



42. Again, if the crowns are parallel discs and the initial 
element of the bucket is radial, and if the water quits the wheel 
radially, and if the velocity of whirl equals the velocity of the 
initial rim, we have 

^ U = Mr? oo% (136) 

as in equation (134). But y 2 will not be the same as in (135). 
To find it we have, neglecting the thickness of the walls of the 
buckets, 

2 n r 1 v l — 2 n r 2 sin y 2 • v 2 
v 1 — Y siii a 
r x go = V cos a 
w = r 2 go tan y 2 
v 2 = r % go ; 

r? tan a 

.-. tan y 2 = ; (137) 

\/r? — r? tan 2 a 

,:U= 8 Q H-\Mw 2 

r? tan 2 a 
= 8QH-\Mr!oS --—• (138) 



r? — r? tan 2 a 



Equations (136) and (138) give 



r x go = 



it/ *9H 

I / r? r? tan 2 a 

y 2 + ^ 

r* — r* tan 2 a 



(139) 



HYDRAULIC MOTOR8. 63 

43. Conclusions.- From an examination of Tables II., III., 
VI., VIL, the following conclusions are drawn: 

1. The maximum theoretical efficiency of the inflow wheel is 
perceptibly larger than that of the outflow, the width of crowns 
and the initial and terminal angles of the buckets being the same, 
One reason for this is due to the flow through the wheel being 
opposed by the centrifugal action, but more particularly to the 
smaller velocity of discharge from the inflow wheel. 

2. Columns (10) in Tables VI. and VII. show that for the 
wheels here considered the loss of energy due to the quitting 
velocity is from 2.2, 5.1 per cent, from the outflow, and from 0.9 
to 1 per cent for the inflow. 

3. The same tables show that in column (2) the efficiency 
is almost constant for the varying conditions here considered, while 
for the outflow there is considerable variation. 

1. One of the most interesting and profitable studies to the 
theorist and practitioner is the effect upon the efficiency due to 
properly proportioning the terminal angle, a, of the guide blade. 
Jt will be observed that all the efficiencies in Tables VI. and VII. 
exceed the corresponding ones in Table II. except the first in 
column (3) of Table II. In Table II. the terminal angle, a, is 
('•distantly 25°, while in Tables VI. and VII. it is less than that 
value, and in the highest efficiencies very much less. 

5. It appears from these tables that the terminal angle, or, 
has frequently been made too large for best efficiency. 

6. That the terminal angle, a, of the guide should be compara- 
tively small for best effect ; for the inflow less than 10°, and that 
theoretically, when the angle is about 7°, the efficiency is some 10 
per cent, greater than when it is 25° in the wheels here considered. 

7. Tables II. and VI. indicate that the initial angle of the 
bucket should exceed 90° for best effect for outflow wheels. 

8. Tables II. and VII. show that the initial angle should be 
less than 90° for best effect on inflow wheels, but that from 60° 
to 120° the efficiency varies scarcely 1 per cent. 



64 HYDRAULIC MOTORS. 

9. The most marked effect in properly proportioning the ter- 
minal angle, «, of the guide is shown when the initial angle of 
the bucket is 150°. In this case the efficiency for the outflow 
when a is 25° is 0.744, Table II., but when a is 13J°, as in Table 
VI., it becomes 0.921. For the inflow, in the former case, it is 
0.752, but when the angle is 3°, as in Table YIL, it becomes 
0.918. 

10. Since the wheels here considered have the same width of 
crowns and the same terminal angle of the bucket, the depths of 
the wheels will be proportional to \ for discharging equal vol- 
umes of water. Tables III., VI., VII. show that the section ~k % in- 
creases as the initial angle of the buckets increases, and that it must 
be greater for the inflow than for the outflow ; hence the depth 
of the wheel must be greater for the inflow for delivering the 
same volume of water. 

11. But the same volume of water delivered by the inflow does 
more work than that of the outflow ; the depths should be as A' 2 , 
divided by the efficiency. Thus in Tables YL and YIL, for 
y = 90°, and for the same heads, II the relative depths should 
be for equal works (0.759 -f- 0.828) -=- (150 -^ 0.920) == 1.67. 

12. In the outflow wheel, column (9), Table VI., shows that for 
the outflow for best effect the direction of the quitting water in 
reference to the earth should be nearly radial (from 76° to 97°), 
but for the inflow wheel the water is thrown forward in quitting 
(column [9] Table YIL). This alone shows that the velocity of 
the rim should somewhat exceed the relative final velocity back- 
ward in the bucket, as shown in columns (4) and (5). 

13. In these tables I have given all the velocities in terms of 
V 2 g A, and the coefficients of this expression will be the part 

of the head which would produce that velocity if the water issued 
freely. In Tables YL and YIL there is only one case, column 
(5) of the former table, where the coefficient exceeds unity, and 
the excess is so small it may be discarded ; and it may be said 
that in a properly proportioned turbine with the conditions here 



HYDRAULIC MOTORS. 65 

given, none of the velocities will equal that due to the head in 
the supply chamber when running at best effect. 

14. The inflow turbine presents the best conditions for con- 
struction for producing a given effect, the only apparent disad- 
vantage being an increased first cost due to an increased depth, 

or an increased diameter for producing a given amount of work. 
The larger efficiency should, however, more than neutralize the 
increased rirst cost. 

Column (3) shows that the efficiency, E, increases as the 
initial angle of the bucket, y iy increases, up to 120°. The maxi- 
mum will be for about 120° with this amount of friction. 



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"The most complete work on the subject ever published."— American 
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"Here is an invaluable book to the practical workman and amateur."— 
London Weekly Times. 

A TREATISE ON TOOTHED GEARING. 

Containing complete instructions of Designing, Drawing, and 
Constructing Spur Wheels, Bevel Wheels, Lantern Gear, 
Screw Gear, Worms, etc., and the proper formation of Tooth 
Profiles. For the use of Machinists, Pattern Makers, Draughts- 
men, Designers, Scientific Schools, etc. With many plates. By 

J. Howard Cromwell. Fourth edition 12mo, cloth, 1 50 

"Mr. Cromwell has accomplished good work in bringing together in this 
volume a great deal of information only to be found by searching many 
works, and by adding the results of his own experience in the field of 
Mechanical Engineering." — American Machinist. 

A TREATISE ON BELTS AND PULLEYS. 

Embracing full explanations of Fundamental Principles ; 
proper Disposition of Pulleys ; Rules for determining widths 
of leather and vulcanized rubber belts, and belts running over 
covered pulleys ; Strength and Proportions of Pulleys, Drums, 
etc. Together with the principles and necessary rules for 
Rope Gearing and transmission of power by means of Metallic 
Cables. By J. Howard Cromwell, Ph. B., author of a Treatise 

on Toothed Gearino- 12mo, cloth, 1 50 

"This is a very complete and comprehensive treatise, and is worthy of 
the attention of all Mechanics who have anything to do with the manage- 
ment of belts and pulleys, etc."— National Car Builder. 

SAW FILING. 

The Art of Saw Filing Scientifically Treated and Explained 
on Philosophical Principles. With explicit directions for 
putting in order all kinds of Saws, from a Jeweler's Saw to a 
Steam Saw-mill. Illustrated by 44 engravings. By H. W. 
Holly. Fifth edition 18mo, cloth, 75 

SAW FILING. 

A Practical Treatise on Filing, Gumming, and Swageing Saws. 

By Robert H. Grimshaw Fully illustrated 1 vol., 16mo, 1 00 

MACHINERY PATTERN MAKING. 

A Discussion of Methods, including Marking and Recording 
Patterns, Printing Press, Slice Valve and Corliss Cylinders ; 
How to Cast Journal-boxes on Frames, Differential Pulleys, 
Fly-wheels, Engine Frames, Spur, Bevel and Worm Gears, 
Key Heads for Motion Rods, Elbows and Tee Pipes, Sweeping 
Straight and Conical Grooved Winding Drums, Large Sheaves 
with Wrought and Cast-iron Arms, 128 full-size Profiles of 



MECHANICS— MACHINERY. 



Gear Teeth of different pitches for Gears of 14 to 800 Teeth, 
with a Table showing at a glance the required diameter of 
Gear for a given number of teeth and pitch, Double Beat, 
Governor, and Plug Valves, Screw Propellor, a chapter on 
items for Pattern Makers, besides a number of valuable end 
useful Tables, etc., etc. 417 illustrations By P S. Dingey, 
Foreman Pattern Maker and Draughtsman. . . .12mo, cloth, $2 00 

" A neat little work that should be not only in the hands of every pattern 
maker, but, read by every foundry foreman and proprietor of foundries 
doing machinery work."— Machinery Moulder's Journal. 

THE BOSTON MACHINIST. 

Being a complete School for the Apprentice as well as the 
advanced Machinist, showing how to make and use every tool 
in every branch of the business ; with a Treatise on Screws and 
Gear-cutting. By "Walter Fitzgerald. Third ed'n. 18mo,cloth, 75 

STEAM HEATING FOR BUILDINGS. 

Or, Hints to Steam Fitters. Being a description of Steam 
Heating Apparatus for Warming and Ventilating Private 
Houses and Large Buildings, with Remarks on Steam, Water, 
and Air in their Relations to Heating. To which are added 
useful miscellaneous tables. By Wm. J. Baldwin. Thirteenth 

edition. With many illustrative plates 12mo, cloth, 2 50 

"Mr. Baldwin has supplied a want long felt for a practical work 
on Heating and Heating Apparatus. 1 '— Sanitary Engineer. 

THE COST OF MANUFACTURES— AND THE AD- 
MINISTRATION OF WORKSHOPS, PUBLIC 
AND PRIVATE. 

A System of Mechanical Book-keeping, based on the Card- 
Catalogue method, dispensing with skilled clerical labor and 
the use of books, by which the cost of manufactures may be 
promptly determined, either in gross or in any detail, as to 
component parts and operations thereon. Comprising a 
simple method of recording all dealings with materials which 
relate to its procurement, expenditure, or possession. Applied, 
with numerous practical illustrations, to the trust, accounta- 
bility for public property, and funds required of the U. S. 
Ordnance Department, with a review of its present practice. 
ByCapt. Henry Metcalfe, U. S. Ordnance Department. Illustra- 
ted with tables, forms of cards, etc., etc. Second ed'n. 8vo,cloth, 5 00 

" I feel sure that by the use of your methods I can determine a cost I hare 
never been able to arrive at." — Ewart Manufacturing Co., Chicago. 

" We tind that it enables us to keep a more accurate record of each piece 
of work. We can locate the responsibility for any delay or omission."— 
Rathbone & Co., Stove Works, Albany. 

WRINKLES AND RECIPES. 

Compiled from the SCIENTIFIC AMERICAN. A collection 
of practical suggestions, processes, and directions, for the 
MECHANIC, ENGINEER, FARMER, and Housekeeper. 
With a COLOR TEMPERING SCALE, and numerous wood 
engravings. By Park Benjamin. Revised by Profs. Thurston 
and Van der Weyde, and Engineers Buel and Rose. Fifth 

revised edition 12mo, cloth, 2 00 

"Hundreds of Trade Secrets and Mechanical Shop Wrinkles.' 1 

JOHN WILEY & SONS, 
53 E. Tenth St., New York. 



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